Euler's Formula and De Moiver’s Theorem

We know about complex numbers(z). They are of the form z=a+ib, where a and b are real numbers and 'i' is the solution of equation x²=-1. No real number can satisfy this equation hence its solution that is 'i' is called an imaginary number. When a  complex exponential is written, it is written as e^iθ.

Euler's formula explains the relationship between complex exponentials and trigonometric functions.

DeMoivers’ theorem is also a theorem used for complex numbers. This theorem is used to raise complex numbers to different powers.

State Euler's Theorem

Euler’s law states that ‘For any real number x, e^ix = cos x + i sin x.

where,e=base of natural logarithm

           i=imaginary unit

          x=angle in radians

This complex exponential function is sometimes denoted cis x ("cosine plus i sine"). The formula is still valid if x is a complex number.

Let z be a non zero complex number; we can write z in the polar form as,

z = r(cos θ + i sin θ) = r e^iθ, where r is the modulus and θ is argument of z.

Multiplying a complex number z with e^iα gives, zei^α = re^iθ × ei^α = rei^(α + θ).The resulting complex number re^i(α+θ) will have the same modulus r and argument (α+θ).

Euler's Identity 

When x=π Euler’s formula evaluates to e^iπ+1=0, which is known as Euler's Identity.

Image to be added soon

Euler's Formula 

Euler's Formula For Cube

Euler's formula is related to the Faces, Edges and vertices of any polyhedron. 

Euler's formula for a cube says that in a cube, the number of vertices minus the number of edges plus the number of faces results in two.

It can be written as 

                              V-E+F=2

Where, V=number of vertices

           E=number of edges

           F=number of faces

It can be proven as,

In a cube, the number of vertices = 8

                number of edges= 12

                number of faces= 6

Putting values into the formula, V-E+F=8-12+6

                                                     =2

Hence proved.

De Moiver's Theorem

State De Moiver's Theorem

It states that for any integer n,

(cos θ + i sin θ)^n = cos (nθ) + i sin (nθ)

We can prove this easily using Euler’s formula as given below,

We know that, (cos θ + i sin θ) = e^iθ

(cos θ + i sin θ)^n = e^i(nθ)

Therefore,

e^i(nθ) = cos (nθ) + i sin (nθ)

Image will be added soon

nth Roots of Unity

If any complex number satisfies the equation zn = 1, it is known as nth root of unity.

An equation of degree n will have n roots as said by the fundamental theory of algebra, there are n values of z which satisfies zn = 1.

To find the values of z, we can write,

1 = cos (2kπ) + i sin (2kπ), —(1) where k can be any integer.

We have,

z^n = 1

z = 1^(1/n)

From (1),

z = [cos (2kπ) + i sin (2kπ)]^(1/n)

By De Moivre’s theorem,

z = [cos (2kπ/n) + i sin (2kπ/n)], where k = 0,1,2,3,……..,n−1

For example; if n = 3, then k = 0,1,2

We know that, z = cos (2kπ/n) + i sin (2kπ/n) = e^i(2kπ/n)

Let ω = cos (2πn) +i sin (2πn) = e^i(2πn)

nth roots of unity are found by,

When k = 0; z = 1

k = 1; z = ω

k = 2; z = ω²

k = n; z = ω^{n − 1}

Therefore, nth roots of unity are 1,ω,ω^2,ω³,…….,ω^{n − 1}

Sum of nth roots of unity is,1 + ω + ω² + ω³ + ⋯ + ω^{n − 1}It is geometric series having first term 1 and common ratio ω.By using sum of n terms of a G.P,1 + ω + ω² + ω³ + ⋯ + ω^{n − 1} = 1 − ω^{n1 − ω}Since ω is nth root of unity, ωn = 1Therefore, 1 + ω + ω² + ω³ + ⋯ + ω^{n − 1} = 0

Cube Roots of Unity:

We know that nth roots of unity are 1,ω,ω2,ω3,…….,ωn − 1.

Therefore, cube roots of unity are 1,ω,ω2 where,

ω = cos (2π/3) + i sin (2π/3) = −1 + √3 i2

ω² = cos(4π/3) + i sin (4π/3) = −1 − √3 i2

Sum of the cube roots of the unity,

1 + ω + ω² = 0

Product of cube roots of the unity,

1 × ω × ω² = ω³ = 1

De Moiver's Theorem Example 

If z = (cosθ + i sinθ ) , show that z^n + 1/ z^n = 2 cos nθ and z^n – [1/ z^n] = 2i sin nθ .

Solution

Let z = (cosθ + i sinθ ) . 

By de Moivre’s theorem ,

z^n = (cosθ + i sinθ )^n = cos nθ + i sin nθ

1/z^n=z^(-n)=cos nθ - i sin nθ

=> z^n+1/z^n = (cos nθ + i sin nθ)+(cos nθ - i sin nθ)

=> z^n+1/z^n = 2cosnθ

Also,=> z^n-1/z^n = (cos nθ + i sin nθ)-(cos nθ - i sin nθ)

=> z^n-1/z^n = 2i sin nθ